Required to

A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the

Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$ . What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

$= 3050.19 \ kg \ m^2$

So the moment of inertia is $3050.19 \ kg \ m^2$ .

4 0

3 years ago

Which spots in this picture show where a WARM air mass would form? *

Svet_ta [14]

Answer:

Spot B & Spot C

Explanation:

They're closer to the equator and get more direct solar radiation, making them more likely to be where a warm air mass would form.

3 0

3 years ago

Read 2 more answers

an object of mass 1 g is hung from a spring and set in oscillatory motion .At t=0 the displacement is 43.75cm and the accelerati

charle [14.2K]

The spring constant is $4.0\cdot 10^{-5} N/m$

Explanation:

For an object in a simple harmonic motion, the acceleration of the object is related to the displacement by

$a=-\omega^2 x$

where

a is the acceleration

$\omega$ is the angular frequency

x is the displacement

The angular frequency is defined as

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Substituting the second equation into the first one, we get

$a=-\frac{k}{m}x$

In this problem we have

m = 1 g = 0.001 kg

And at t=0,

x = 43.75 cm

a = -1.754 cm/s

Therefore, we can re-arrange the equation above to find the spring constant:

$k=-\frac{ma}{x}=-\frac{(0.001)(-1.754)}{43.75}=4.0\cdot 10^{-5} N/m$

#LearnwithBrainly

6 0

3 years ago

Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and b

Alona [7]

Answer:

E. The force on A is exactly equal to the force on B.

Explanation:

The force between two charges is given by

$F=\dfrac{kq_1q_2}{r^2}$

where

$q_1$ = Charge on particle 1

$q_2$ = Charge on particle 2

r = Distance between the charges

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

$F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N$

This force will be exerted on both the charges equally.

So, The force on A is exactly equal to the force on B

6 0

3 years ago

Complete the following:

masha68 [24]

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>

If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

brainly.com/question/13095658

#SPJ1

8 0

2 years ago